If tanθ=7/24,where 180<θ<270,fine the values of sinθ and cosθ.
我計黎計去都係計到sinθ=-24/25 cosθ=-7/25
但個答案係岩岩調轉...點解既?
注意:依條問題唔係用黎求做功課,唔該果啲yahoo管理員唔好一收到啲無聊檢舉就睇都唔睇亂咁del
F.4數學問題(三角學)
2010-04-09 10:28 pm
回答 (2)
2010-04-09 11:57 pm
✔ 最佳答案
tanθ=7/24 and180<θ<270, θ lies in quadrant III,
Let x be the hypotenuse,
x^2 = 24^2 + 7^2 (Pythagoras' Theorem)
x = 25
sinθ = opposite side / hypotenuse
= -7/25
cosθ = adjacent side / hypotenuse
= -24/25
2010-04-09 15:58:59 補充：
Moreover, only tanθ is positive in quadrant III.
2010-04-09 10:49 pm
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因 180<θ<270 , 即 θ 在第三象限 , 此時只有 tanθ 是正 , sinθ and cosθ 都是負。
由恒等式 sin^2 θ + cos^2 θ = 1 ,
得 sin^2 θ = 1 - cos^2 θ ......(1)
又 tan θ = sin θ / cos θ , 兩邊平方 :
tan^2 θ = sin^2 θ / cos^2 θ , 把(1)代入 :
(7/24)^2 = (1 - cos^2 θ) / cos^2 θ
(49/576)cos^2 θ = 1 - cos^2 θ
(625/576) cos^2 θ = 1
cos^2 θ = 576/625
cos θ = - 24/25 or 24/25 (rejected since 180<θ<270 , cos θ is always negative)
So sin^2 θ = 1 - cos^2 θ = 1 - 576/625 = 49/625
sin θ = - 7/25 or 7/25 (rejected since 180<θ<270 , sin θ is always negative)
因 180<θ<270 , 即 θ 在第三象限 , 此時只有 tanθ 是正 , sinθ and cosθ 都是負。
由恒等式 sin^2 θ + cos^2 θ = 1 ,
得 sin^2 θ = 1 - cos^2 θ ......(1)
又 tan θ = sin θ / cos θ , 兩邊平方 :
tan^2 θ = sin^2 θ / cos^2 θ , 把(1)代入 :
(7/24)^2 = (1 - cos^2 θ) / cos^2 θ
(49/576)cos^2 θ = 1 - cos^2 θ
(625/576) cos^2 θ = 1
cos^2 θ = 576/625
cos θ = - 24/25 or 24/25 (rejected since 180<θ<270 , cos θ is always negative)
So sin^2 θ = 1 - cos^2 θ = 1 - 576/625 = 49/625
sin θ = - 7/25 or 7/25 (rejected since 180<θ<270 , sin θ is always negative)
收錄日期: 2021-04-21 22:12:11
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100409000051KK00786